ee202 - ee math ii jitkomut songsiri 12. residues and its...

EE202 - EE MATH II Jitkomut Songsiri

12. Residues and Its Applications

• isolated singular points

• residues

• Cauchy’s residue theorem

• applications of residues

12-1

Isolated singular points

z0 is called a singular point of f if

• f fails to be analytic at z0

• but f is analytic at some point in every neighborhood of z0

a singular point z0 is said to be isolated if f is analytic in some punctured disk

0 < |z − z0| < ε

centered at z0 (also called a deleted neighborhood of z0)

example: f(z) = 1/(z2(z2 + 1)) has the three isolated singular points at

z = 0, z = ±j

Residues and Its Applications 12-2

Non-isolated singular points

example: the function1

sin(π/z)has the singular points

z = 0, z =1

n, (n = ±1,±2, . . .)

• each singular point except z = 0 is isolated

• 0 is nonisolated since every punctured disk of 0 contains other singularities

• for any ε > 0, we can find a positive integer n such that n > 1/ε

• this means z = 1/n always lies in the punctured disk 0 < |z| < ε


Residues

assumption: z0 is an isolated singular point of f , e.g.,

there exists a punctured disk 0 < |z − z0| < r0 throughout which f is analytic

consequently, f has a Laurent series representation

f(z) =

∞∑n=0

an(z − z0)n +b1

z − z0+ · · ·+ bn

(z − z0)n+ · · · , (0 < |z − z0| < r0)

let C be any positively oriented simple closed contour lying in the disk

0 < |z − z0| < r0

the coefficient bn of the Laurent series is given by

bn =1

j2π

∫C

f(z)

(z − z0)−n+1dz, (n = 1, 2, . . .)


the coefficient of 1/(z − z0) in the Laurent expansion is obtained by∫C

f(z)dz = j2πb1

b1 is called the residue of f at the isolated singular point z0, denoted by

b1 = Resz=z0

f(z)

this allows us to write

∫C

f(z)dz = j2π Resz=z0

f(z)

which provides a powerful method for evaluating integrals around a contour


example: find∫Ce1/z

2dz when C is the positive oriented circle |z| = 1

1/z2 is analytic everywhere except z = 0; 0 is an isolated singular point

the Laurent series expansion of f is

f(z) = e1/z2

= 1 +1

z2+

1

2!z4+

1

3!z6+ · · · (0 < |z| <∞)

the residue of f at z = 0 is zero (b1 = 0), so the integral is zero

remark: the analyticity of f within and on C is a sufficient condition for∫Cf(z)dz to be zero; however, it is not a necessary condition


example: compute∫C

1

z(z + 2)3dz where C is circle |z + 2| = 1

f has the isolated singular points at 0 and −2

choose an annulus domain: 0 < |z + 2| < 2

on which f is analytic and contains C

f has a Laurent series on this domain and is given by

f(z) =1

(z + 2− 2)(z + 2)3= −1

2· 1

1− (z + 2)/2· 1

(z + 2)3

= − 1

2(z + 2)3

∞∑n=0

(z + 2)n

2n= −

∞∑n=0

(z + 2)n−3

2n+1, (0 < |z + 2| < 2)

the residue of f at z = −2 is −1/23 which is obtained when n = 2

therefore, the integral is j2π(−1/23) = −jπ/4 (check with the Cauchy formula)


Cauchy’s residue theorem

let C be a positively oriented simple closed contour

Theorem: if f is analytic inside and on C except for a finite number of singularpoints z1, z2, . . . , zn inside C, then

∫C

f(z)dz = j2π

n∑k=1

Resz=zk

f(z)

Proof.

• since zk’s are isolated points, we can find smallcircles Ck’s that are mutually disjoint

• f is analytic on a multiply connected domain

• from the Cauchy-Goursat theorem:∫Cf(z)dz =

∑nk=1

∫Ckf(z)dz


example: use the Cauchy residue theorem to evaluate the integral∫C

3(z + 1)

z(z − 1)(z − 3)dz, C is the circle |z| = 2, in counterclockwise

C encloses the two singular points of the integrand, so

I =

∫C

f(z)dz =

∫C

3(z + 1)

z(z − 1)(z − 3)dz = j2π

[Resz=0

f(z) + Resz=1

f(z)]

• calculate Resz=0 f(z) via the Laurent series of f in 0 < |z| < 1

• calculate Resz=1 f(z) via the Laurent series of f in 0 < |z − 1| < 1


rewrite f(z) =1

z−

3

z − 1+

2

z − 3

• the Laurent series of f in 0 < |z| < 1

f(z) =1

z+

3

1− z− 2

3(1− z/3)=

1

z+3(1+z+z2+. . .)−2

3(1+(z/3)+(z/3)2+. . .)

the residue of f at 0 is the coefficient of 1/z, so Resz=0 f(z) = 1

• the Laurent series of f in 0 < |z − 1| < 1

f(z) =1

1 + z − 1− 3

z − 1− 1

1− (z − 1)/2

= 1− (z − 1) + (z − 1)2 + . . .− 3

z − 1−

(1 +

z − 1

2+

(z − 1

2

)2

+ . . .

)

the residue of f at 1 is the coefficient of 1/(z − 1), so Resz=0 f(z) = −3


therefore, I = j2π(1− 3) = −j4π

alternatively, we can compute the integral from the Cauchy integral formula

I =

∫C

(1

z−

3

z − 1+

2

z − 3

)dz

= j2π(1− 3 + 0) = −j4π


Residue at infinity

f is said to have an isolated point at z0 =∞ if

there exists R > 0 such that f is analytic for R < |z| <∞

singular points of

C is a positive oriented simple closed contour

Theorem: if f is analytic everywhere except for a finite number of singularpoints interior to C, then∫

C

f(z)dz = j2πResz=0

[1

z2f

(1

z

)](see a proof on section 71, Churchill)


example: find I =∫C

z − 3

z(z − 1)dz, C is the circle |z| = 2 (counterclockwise)

I = j2πResz=0

[(1/z2)f(1/z)

]= j2πRes

z=0

[1− 3z

z(1− z)

], j2πRes

z=0g(z)

find the residue via the Laurent series of g in 0 < |z| < 1

write g(z) =

(1

z− 3

)(1 + z + z2 + · · · ) =⇒ Res

z=0g(z) = 1

compare the integral with other methods .

• Cauchy integral formula (write the partial fraction of f)

• Cauchy residue theorem (have to find two residues; hence two Laurent series)


Principal part

f has an isolated singular point at z0, so f has a Laurent seires

f(z) =

∞∑n=0

an(z − z0)n +b1

(z − z0)+

b2(z − z0)2

+ · · ·+ bn(z − z0)n

+ · · ·

in a punctured disk 0 < |z − z0| < R

the portion of the series that involves negative powers of z − z0

b1(z − z0)

+b2

(z − z0)2+ · · ·+ bn

(z − z0)n+ · · ·

is called the principal part of f


Types of isolated singular points

three possible types of the principal part of f

• no principal part

f(z) = cos z = 1− z2

2!+z4

4!+ · · · , (0 < |z| <∞)

• finite number of terms in the principal part

f(z) =1

z2(1 + z)=

1

z2− 1

z+ 1− z + z2 + · · · , (0 < |z| < 1)

• infinite number of terms in the principal part

f(z) = e1/z = 1 +1

z+

1

2!z2+

1

3!z3+ · · · , (0 < |z| <∞)


classify the number of terms in the principal part in a general form

• none: z0 is called a removable singular point

f(z) =

∞∑n=0

an(z − z0)n

• finite (m terms): z0 is called a pole of order m

f(z) =

∞∑n=0

an(z − z0)n +b1

z − z0+

b2(z − z0)2

+ · · ·+ bm(z − z0)m

• infinite: z0 is said to be an essential singular point of f

f(z) =

∞∑n=0

an(z − z0)n +b1

z − z0+

b2(z − z0)2

+ · · ·+ bn(z − z0)n

+ · · ·


examples:

f1(z) = cos z = 1− z2

2!+z4

4!+ · · ·

f2(z) =3

(z − 1)(z − 2)= −

(1

z − 2+ 1 + (z − 2) + (z − 2)3 + · · ·

)f3(z) =

1

z2(1 + z)=

1

z2− 1

z+ 1− z + z2 + · · ·

f4(z) = e1/z = 1 +1

z+

1

2!z2+

1

3!z3+ · · ·

• 0 is a removeable singular point of f1

• 2 is a pole of order 1 (or simple pole) of f2

• 0 is a pole of order 2 (or double pole) of f3

• 0 is an essential singular point of f4

note: for f2, f3 we can determine the pole/order from the denominator of f


Characterization of poles

an isolated singular point z0 of a function f is a pole of order m if and only if

f(z) =φ(z)

(z − z0)m

where φ(z) is analytic and nonzero at z0

Proof. since φ is analytic at z0, it has Taylor series about z = z0

φ(z) = φ(z0) + · · ·+ φ(m−1)(z0)(z − z0)m−1

(m− 1)!+

∞∑k=m

φ(k)(z0)(z − z0)k

k!

f(z) =φ(z0)

(z − z0)m+ · · ·+ φ(m−1)(z0)

(m− 1)!(z − z0)+

∞∑k=m

φ(k)(z0)(z − z0)k−m

k!

f has a pole at z0 of order m when φ is nonzero at z0


Residue formula

if f has a pole of order m at z0 then

Resz=z0

f(z) =1

(m− 1)!limz→z0

dm−1

dzm−1(z − z0)mf(z)

Proof. if f has a pole of order m, its Laurent series can be expressed as

f(z) =

∞∑n=0

an(z − z0)n +b1

(z − z0)+

b2(z − z0)2

+ · · ·+ bm(z − z0)m

(z − z0)mf(z) =

∞∑n=0

an(z − z0)m+n + b1(z − z0)m−1 + b2(z − z0)m−2 + · · ·+ bm

to obtain b1, we take the (m− 1)th derivative and take the limit z → z0


example 1: find Resz=0 f(z) and Resz=2 f(z) where f(z) =(z + 1)

z2(z − 2)

Resz=0

f(z) = limz→0

d

dz

(z + 1

z − 2

)= −3/4 (0 is a double pole of f)

Resz=2

f(z) = limz→2

z + 1

z2= 3/4

example 2: find Resz=0 g(z) where g(z) =z + 1

1− 2z

g is analytic at 0 (0 is a removable singular point of g), so Resz=0 g(z) = 0

check . apply the results from the above two examples to compute∫C

(z + 1)

z2(z − 2)dz, C is the circle |z| = 3 (counterclockwise)

by using the Cauchy residue theorem and the formula on page 12-12


sometimes the pole order cannot be readily determined

example 3: find Resz=0 f(z) where f(z) =sinh z

z4

use the Maclaurin series of sinh z

f(z) =1

z4·(z +

z3

3!+z5

5!+ · · ·

)=

(1

z3+

1

3!z+z

5!+ · · ·

)0 is the third-order pole with residue 1/3!

here we determine the residue at z = 0 from its definition (the coeff. of 1/z )

no need to use the residue formula on page 12-19


L’Hopital’s rule for complex functions

let f(z) and g(z) be analytic in a region containing z0 and

• f(z0) = g(z0) = 0

• g′(z0) 6= 0

then the L’Hopital’s rule states that

limz→z0

f(z)

g(z)=f ′(z0)

g′(z0)

in case f ′(z0) = g′(z0) = 0, the rule may be extended

example: limz→0sin zz = 1


when the pole order (m) is unknown, we can

• assume m = 1, 2, 3, . . .

• find the corresponding residues until we find the first finite value

example 4: find Resz=0 f(z) where f(z) =1 + z

1− cos z

• assume m = 1

Resz=0

f(z) = limz→0

z(1 + z)

1− cos z= 0/0 = lim

z→0

1 + 2z

sin z= 1/0 =∞ =⇒ (not 1st order)

• assume m = 2

Resz=0

f(z) = limz→0

d

dz

(z2(1 + z)

1− cos z

)= 2 (finite) =⇒ 0 is a double pole

note: use L’Hopital’s rule to compute the limit


Summary

many ways to compute a contour integral (∫Cf(z)dz)

• parametrize the path (feasible when C is easily described)

• use the principle of deformation of paths (if f is analytic in the regionbetween the two contours)

• use the Cauchy integral formula (typically requires the partial fraction of f)

• use the Cauchy’s residue theorem on page 12-8 (requires the residues atsingular points enclosed by C)

• use the theorem of the residue at infinity on page 12-12 (find one residue at 0)

to find the residue of f at z0

• read from the coeff of 1/(z − z0) in the Laurent series of f

• apply the residue formula on page 12-19


Application of the residue theorem

• improper integrals

• improper integrals from Fourier series

• inversion of Laplace transforms

• integrals involving sines and cosines

ingredients: residue theorem, upper bound of contour integral, Jordan inequality


Improper integrals

let’s first consider a well-known improper integral

I =

∫ ∞−∞

dx

1 + x2= π

of course, this can be evaluated using the inverse tangent function

we will derive this kind of integral by means of contour integration

some poles of the integrand lie in the upper half plane

let CR be a semicircular contour with radius R→∞

∫ R

−Rf(x)dx+

∫CR

f(z)dz = j2π∑k

Resz=zk

f(z)

and show that∫CRf(z)dz → 0 as R→∞


Theorem: if all of the following assumptions hold

1. f(z) is analytic in the upper half plane except at a finite number of poles

2. none of the poles of f(z) lies on the real axis

3. |f(z)| ≤M

Rkwhen z = Rejθ; M is a constant and k > 1

then the real improper integral can be evaluated by a contour integration, and∫ ∞−∞

f(x)dx = j2π

[sum of the residues of f(z) at the poleswhich lie in the upper half plane

]

• assumption 2: f is analytic on C1

• assumption 3:∫CRf(z)dz → 0 as R→∞


Proof. consider a semicircular contour with radius R large enough to include allthe poles of f(z) that lie in the upper half plane

• from the Cauchy’s residue theorem∫C1∪CR

f(z)dz = j2π[∑

Res f(z) at all poles within C1 ∪ CR]

(to apply this, f(z) cannot have singular points on C1, i.e., the real axis)

• the integral along the real axis is our desired integral

limR→∞

∫ R

−Rf(x)dx+ lim

R→∞

∫CR

f(z)dz = limR→∞

∫C1∪CR

f(z)dz

• hence, it suffices to show that

limR→∞

∫CR

f(z)dz = 0 by using |f(z)| ≤M/Rk, where k > 1


(continue the proof of applying residue theorem)

• apply the modulus of the integral and use |f(z)| ≤M/Rk∣∣∣∣∫CR

f(z)dz

∣∣∣∣ ≤ M

Rk· length of CR =

MπR

Rk

hence, limR→∞∫CRf(z)dz = 0 if k > 1

remark: an example of f(z) that satisfies all the conditions in page 12-27

f(x) =p(x)

q(x), p and q are polynomials

q(x) has no real roots and deg q(x) ≥ deg p(x) + 2

(relative degree of f is greater than or equal to 2)


example: show that ∫CR

f(z)dz = 0

as R→∞ where CR is the arc z = Rejθ, 0 ≤ θ ≤ π

• f(z) = (z + 2)/(z3 + 1) (relative degree of f is 2)

|z + 2| ≤ |z|+ 2 = R+ 2, |z3 + 1| ≥ ||z3| − 1| = |R3 − 1|

hence, |f(z)| ≤ R+2R3−1 and apply the modulus of the integral

∣∣∣∣∫C

f(z)dz

∣∣∣∣ ≤ ∫C

|f(z)|dz ≤ R+ 2

R3 − 1· πR = π ·

1 + 2R2

R− 1R2

the upper bound tends to zero as R→∞


• f(z) = 1/(z2 + 2z + 2)

z2 + 2z + 2 = (z − (1 + j))(z − (1− j)) , (z − z0)(z − z0)

hence, |z − z0| ≥ ||z| − |z0|| = R− |1 + j| = R−√

2 and similarly,

|z − z0| ≥ ||z| − |z0|| = R−√

2

then it follows that

|z2 + 2z + 2| ≥ (R−√

2)2 ⇒ |f(z)| ≤ 1

(R−√

2)2∣∣∣∣∫C

f(z)dz

∣∣∣∣ ≤ ∫C

|f(z)|dz ≤ 1

(R−√

2)2· πR =

π

(1−√2R )2

the upper bound tends to zero as R→∞


example: compute I =

∫ ∞−∞

dx

1 + x2

• define f(z) =1

1 + z2and create a contour C = C1 ∪ CR as on page 12-26

• relative degree of f is 2, so∫CRf(z)dz = 0 as R→∞

• f(z) has poles at z = j and z = −j (no poles on the real axis)

• only the pole z = j lies in the upper half plane

• by the residue’s theorem

j2π ·∑

Resz=zk

f(z) =

∮C

f(z)dz =

∫ R

−Rf(x)dx︸︷︷︸

=I as R→∞

+

∫CR

f(z)dz︸︷︷︸=0 as R→∞

I = j2πResz=j

f(z) = j2π limz→j

(z − j)f(z) = π


example: compute

I =

∫ ∞−∞

x2

(x2 + a2)(x2 + b2)dx

• define f(z) =z2

(z2 + a2)(z2 + b2)and create C = C1 ∪ CR as on page 12-26


• f(z) has poles at z = ±ja and z = ±jb (no poles on the real axis)

• only the poles z = ja and z = jb lie in the upper half plane

• by the residue’s theorem

j2π ·∑

Resz=zk

f(z) =

∮C

f(z)dz =

∫ R


=I as R→∞

+

∫CR

f(z)dz︸︷︷︸=0 as R→∞

I = j2π

[Resz=ja

f(z) + Resz=jb

f(z)

]= j2π

[a

j2(a2 − b2)+

b

j2(b2 − a2)

]=

π

a+ b


Applications of residue theorem






Improper integrals from Fourier analysis

we can use residue theory to evaluate improper integrals of the form∫ ∞−∞

f(x) sinmx dx,

∫ ∞−∞

f(x) cosmx dx, or

∫ ∞−∞

ejmxf(x) dx

we note that ejmz is analytic everywhere, moreover

|ejmz| = ejm(x+jy) = e−my < 1 for all y in the upper half plane

therefore, if |f(z)| ≤M/Rk with k > 1, then so is |ejmzf(z)|

hence, if f(z) satisfies the conditions in page 12-27 then∫ ∞−∞

ejmxf(x)dx = j2π

[sum of the residues of ejmzf(z) at the poleswhich lie in the upper half plane

]


denote

S =

[sum of the residues of ejmzf(z) at the poleswhich lie in the upper half plane

]and note that S can be complex

by comparing the real and imaginary part of the integral∫ ∞−∞

ejmxf(x)dx =

∫ ∞−∞

(cosmx+ j sinmx)f(x)dx = j2πS

we have ∫ ∞−∞

cosmxf(x) dx = Re(j2πS) = −2π · ImS∫ ∞−∞

sinmxf(x) dx = Im(j2πS) = 2π · ReS



∫ ∞−∞

cosmx dx

1 + x2

• define f(z) = ejmz

1+z2and create C = C1 ∪ CR as on page 12-26


• f has poles at z = j and z = −j (no poles on the real axis)

• the pole z = j lies in the upper half plane

• by residue’s theorem

j2π ·∑

Resz=zk

f(z) =

∮C

f(z)dz =

∫ R


=I as R→∞

+

∫CR

f(z)dz︸︷︷︸=0 as R→∞


• therefore, ∫ ∞−∞

ejmx

1 + x2dx = j2πRes

z=j

ejmz

1 + z2

= j2π limz→j

(z − j)ejmz

1 + z2= πe−m

• our desired integral can be obtained by∫ ∞−∞

cosmx dx

1 + x2= Re(πe−m) = πe−m,∫ ∞

−∞

sinmx dx

1 + x2= Im(πe−m) = 0


Summary of improper integrals

the examples of f we have seen so far are in the form of

f(x) =p(x)

q(x)

where p, q are polynomials and deg p(x) ≥ deg q(x) + 2


the assumption on the degrees of p, q is sufficient to guarantee that∫CR

f(z)ejazdz = 0 (a > 0)

as R→∞ where CR is the arc z = Rejθ, 0 ≤ θ ≤ π

we can relax this assumption to consider function f such as

z

z2 + 2z + 2,

1

z + 1(relative degree is 1)

and obtain the same result by making use of Jordan’s inequality


Jordan inequality

for R > 0, ∫ π

0

e−R sin θdθ <π

R

Proof.

sin θ ≥ 2θ/π, 0 ≤ θ ≤ π

2

e−R sin θ ≤ e−2Rθ/π, R > 0, 0 ≤ θ ≤ π

2∫ π/2

0

e−R sin θdθ ≤ π

2R

the last line is another form of the Jordan inequality

because the graph of y = sin θ is symmetric about the line θ = π/2


example: let f(z) =z

z2 + 2z + 2show that

∫CRf(z)ejazdz = 0 for a > 0 as

R→∞

• first note that |ejaz| = |eja(x+jy)| = |ejax · e−ay| = e−ay < 1 (since a > 0)

• similar to page 12-32, we see that |f(z)| ≤ R/(R−√

2)2 ,MR and

∣∣∣∣∫CR

f(z)ejazdz

∣∣∣∣ ≤ ∫CR

R

(R−√

2)2· πR =

π

(1−√2R )2

which does not tend to zero as R→∞

• however, for z that lies on CR, i.e., z = Rejθ

f(z)ejaz = f(z)ejaRejθ

= f(z)ejaR(cos θ+j sin θ) = f(z)e−aR sin θ · ejaR cos θ


• if we find an upper bound of the integral, and use Jordan’s inequality:∣∣∣∣∫CR

f(z)ejazdz

∣∣∣∣ =

∣∣∣∣∫ π

0

f(z)e−aR sin θ · ejaR cos θjRejθdθ

∣∣∣∣≤

∫ π

0

∣∣f(z)e−aR sin θ · ejaR cos θjRejθ∣∣ dθ

= RMR

∫ π

0

e−aR sin θdθ

<πMR

a

the final term approach 0 as R→∞ because MR → 0

conclusion: then we can apply the residue’s theorem to integrals like∫ ∞−∞

x cos(ax)

x2 + 2x+ 2dx


Inversion of Laplace transforms

recall the definitions:

F (s) , L[f(t)] ,∫ ∞0

f(t)e−stdt

f(t) = L−1[F (s)] =1

j2π

∫ a+j∞

a−j∞F (s)estds

Theorem: suppose F (s) is analytic everywhere except at the poles

p1, p2, . . . , pn,

all of which lie to the left of the vertical line Re(s) = a (a convergence factor)

if |F (s)| ≤MR and MR → 0 as s→∞ through the half plane Re(s) ≤ a then

L−1[F (s)] =

n∑i=1

Ress=pi

F (s)est


Proof sketch.

parametrize C1 and C2 by

C1 = {z | z = a+ jy, −R ≤ y ≤ R }

C2 =

{z | z = a+Rejθ,

π

2≤ θ ≤ 3π

2

}

1. create a huge semicircle that is large enough to contain all the poles of F (s)

2. apply the Cauchy’s residue theorem to conclude that∫C1

estF (s)ds = j2π

n∑k=1

Ress=pk

[estF (s)]−∫C2

estF (s)ds

3. prove that the integral along C2 is zero when the circle radius goes to ∞


choose a and R: choose the center and radius of the circle

• a > 0 is so large that all the poles of F (s) lie to the left of C1

a > maxk=1,2,...,n

Re(pk)

• R > 0 is large enough so that all poles of F (s) are enclosed by the semicircle

if the maximum modulus of p1, p2, . . . , pn is R0 then

∀k, |pk − a| ≤ |pk|+ a ≤ R0 + a =⇒ pick R > R0 + a

C1 = {z | z = a+ jy, −R ≤ y ≤ R }

C2 =

{z | z = a+Rejθ,

π

2≤ θ ≤ 3π

2

}


integral along C2 is zero

C1 = {z | z = a+ jy, −R ≤ y ≤ R }

C2 =

{z | z = a+Rejθ,

π

2≤ θ ≤ 3π

2

}

• for s = a+Rejθ and ds = jRejθdθ, the integral becomes

∣∣∣∣∫C2

estF (s)ds

∣∣∣∣ =

∣∣∣∣∣∫ 3π/2

π/2

eat · eRt cos θ+jRt sin θF (a+Rejθ)Rjejθdθ

∣∣∣∣∣Residues and Its Applications 12-49

• apply the modolus of the integral∣∣∣∣∫C2

estF (s)ds

∣∣∣∣ ≤ ∫ 3π/2

π/2

∣∣eateRt cos θ · ejRt sin θF (a+Rejθ)Rjejθ∣∣ dθ

• since |F (s)| ≤MR for s that lies on C2∣∣∣∣∫C2

estF (s)ds

∣∣∣∣ ≤MRReat

∫ 3π/2

π/2

eRt cos θ dθ

• make change of variable φ = θ − π/2 and apply the Jordan inequality∣∣∣∣∫C2

estF (s)ds

∣∣∣∣ ≤MRReat

∫ π

0

e−Rt sinφ dφ︸︷︷︸<π/Rt

<πMRe

at

t

the last term approaches zero as R→∞ because MR → 0 (by assumption)


example: find L−1[F (s)] where F (s) =s

(s2 + c2)2and c > 0

C2 ={z | z = a+Rejθ, π

2 ≤ θ ≤3π2

}poles of F (s) are s = ±jc so we choose a > 0

the semicircle must enclose all the pole

so we have R > a+ c

first we verifty that |F (s)| ≤MR and MR → 0 as s→∞ for s on C2

we note that |s| = |a+Rejθ| ≤ a+R and |s| ≥ |a−R| = R− a

since |s2 + c2| ≥ ||s|2 − c2| ≥ (R− a)2 − c2 > 0, then

|F (s)| = |s||s2 + c2|2

≤ (R+ a)

[(R− a)2 − c2]2,MR → 0 as R→∞


therefore, we can apply the theorem on page 12-46

L−1[F (s)] =∑

Ress=sk

[estF (s)] = Ress=jc

sest

(s2 + c2)2+ Ress=−jc

sest

(s2 + c2)2

poles of F (s) are s = ±jc (double poles)

Ress=jc

estF (s) = lims→jc

d

ds

[sest

(s+ jc)2

]=

[est(1 + ts)

(s+ jc)2− 2sest

(s+ jc)3

]s=jc

=tejct

j4c

Ress=−jc

estF (s) = lims→−jc

d

ds

[sest

(s− jc)2

]=

[est(1 + ts)

(s− jc)2− 2sest

(s− jc)3

]s=−jc

= −te−jct

j4c

hence L−1[F (s)] = t4jc(e

jct − e−jct) =t sin ct

2c


example: find L−1[F (s)] where F (s) =1

(s+ a)2 + b2

F (s) has poles at s = −a± jb (simple poles)

L−1[F (s)] = Ress=−a+jb

estF (s) + Ress=−a−jb

estF (s)

(provided that |F (s)| ≤MR and MR → 0 as s→∞ on C2 ... please check .)

Ress=−a+jb

= lims=−a+jb

est

s+ a+ jb=e(−a+jb)t

j2b

Ress=−a−jb

= lims=−a−jb

est

s+ a− jb=e(−a−jb)t

−j2b

hence, L−1[F (s)] =e−at(ejbt − e−jbt)

2jb=e−at sin(bt)

b


Definite integrals involving sines and cosines

we consider a problem of evaluating definite integrals of the form

∫ 2π

0

F (sin θ, cos θ)dθ

since θ varies from 0 to 2π, we can let θ be an argument of a point z

z = ejθ (0 ≤ θ ≤ 2π)

this describe a positively oriented circle C centered at the origin

make the substitutions

sin θ =z − z−1

j2, cos θ =

z + z−1

2, dθ =

dz

jz


this will transform the integral into the contour integral∫C

F

(z − z−1

j2,z + z−1

2

)dz

jz

• the integrand becomes a function of z

• if the integrand reduces to a rational function of z, we can apply the Cauchy’sresidue theorem

example:∫ 2π

0

dθ

5 + 4 sin θ=

∫C

1

5 + 4(z−z−1)2j

dz

jz=

∫C

dz

2z2 + j5z − 2,∫C

g(z)dz

=

∫C

dz

2(z + 2j)(z + j/2)= j2π

(Res

z=−j/2g(z)

)= 2π/3

where C is the positively oriented circle |z| = 1


the above idea can be summarized in the following theorem

Theorem: if F (cos θ, sin θ) is a rational function of cos θ and sin θ which isfinite on the closed interval 0 ≤ θ ≤ 2π, and if f is the function obtained fromF (·, ·) by the substitutions

cos θ =z + z−1

2, sin θ =

z − z−1

j2

then ∫ 2π

C

F (cos θ, sin θ) dθ = j2π

(∑k

Resz=zk

f(z)

jz

)

where the summation takes over all zk’s that lie within the circle |z| = 1



∫ 2π

0

cos 2θ

1− 2a cos θ + a2dθ, −1 < a < 1

make change of variables

• cos 2θ =ej2θ + e−j2θ

2=z2 + z−2

2=z4 + 1

2z2

• 1− 2a cos θ + a2 = 1− 2a(z + z−1)/2 + a2 = −az2 − (a2 + 1)z + a

z

we have∫ 2π

0F (θ)dθ =

∫Cf(z)jz dz ,

∫Cg(z)dz where

g(z) = −(z4 + 1)z

jz · 2z2(az2 − (a2 + 1)z + a)=

(z4 + 1)

j2z2(1− az)(z − a)

we see that only the poles z = 0 and z = a lie inside the unit circle C


therefore, the integral becomes

I =

∫C

g(z)dz = j2π(

Resz=0

g(z) + Resz=a

g(z))

• note that z = 0 is a double pole of g(z), so

Resz=0

g(z) = limz=0

d

dz(z2g(z)) = − 1

j2· a

2 + 1

a2

• Resz=a

g(z) = limz=a

(z − a)g(z) =1

j2· a4 + 1

a2(1− a2)

hence, I =2πa2

1− a2


References

Chapter 6-7 in

J. W. Brown and R. V. Churchill, Complex Variables and Applications, 8thedition, McGraw-Hill, 2009

Chapter 7 in

T. W. Gamelin, Complex Analysis, Springer, 2001

Chapter 22 in

M. Dejnakarin, Mathematics for Electrical Engineering, CU Press, 2006


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